3.290 \(\int \frac{x^2 (a+b x^2+c x^4)}{(d+e x^2)^3} \, dx\)

Optimal. Leaf size=124 \[ \frac{x \left (9 c d^2-e (5 b d-a e)\right )}{8 d e^3 \left (d+e x^2\right )}-\frac{x \left (a e^2-b d e+c d^2\right )}{4 e^3 \left (d+e x^2\right )^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (15 c d^2-e (a e+3 b d)\right )}{8 d^{3/2} e^{7/2}}+\frac{c x}{e^3} \]

[Out]

(c*x)/e^3 - ((c*d^2 - b*d*e + a*e^2)*x)/(4*e^3*(d + e*x^2)^2) + ((9*c*d^2 - e*(5*b*d - a*e))*x)/(8*d*e^3*(d +
e*x^2)) - ((15*c*d^2 - e*(3*b*d + a*e))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(3/2)*e^(7/2))

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Rubi [A]  time = 0.137904, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {1257, 1157, 388, 205} \[ \frac{x \left (9 c d^2-e (5 b d-a e)\right )}{8 d e^3 \left (d+e x^2\right )}-\frac{x \left (a e^2-b d e+c d^2\right )}{4 e^3 \left (d+e x^2\right )^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (15 c d^2-e (a e+3 b d)\right )}{8 d^{3/2} e^{7/2}}+\frac{c x}{e^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*x^2 + c*x^4))/(d + e*x^2)^3,x]

[Out]

(c*x)/e^3 - ((c*d^2 - b*d*e + a*e^2)*x)/(4*e^3*(d + e*x^2)^2) + ((9*c*d^2 - e*(5*b*d - a*e))*x)/(8*d*e^3*(d +
e*x^2)) - ((15*c*d^2 - e*(3*b*d + a*e))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(3/2)*e^(7/2))

Rule 1257

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^
(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[1/(2*e^(2*p +
m/2)*(q + 1)), Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4
)^p - (-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x], x] /; FreeQ[{a, b
, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{x^2 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx &=-\frac{\left (c d^2-b d e+a e^2\right ) x}{4 e^3 \left (d+e x^2\right )^2}-\frac{\int \frac{-c d^2+b d e-a e^2+4 e (c d-b e) x^2-4 c e^2 x^4}{\left (d+e x^2\right )^2} \, dx}{4 e^3}\\ &=-\frac{\left (c d^2-b d e+a e^2\right ) x}{4 e^3 \left (d+e x^2\right )^2}+\frac{\left (9 c d^2-e (5 b d-a e)\right ) x}{8 d e^3 \left (d+e x^2\right )}+\frac{\int \frac{-7 c d^2+e (3 b d+a e)+8 c d e x^2}{d+e x^2} \, dx}{8 d e^3}\\ &=\frac{c x}{e^3}-\frac{\left (c d^2-b d e+a e^2\right ) x}{4 e^3 \left (d+e x^2\right )^2}+\frac{\left (9 c d^2-e (5 b d-a e)\right ) x}{8 d e^3 \left (d+e x^2\right )}-\frac{\left (15 c d^2-e (3 b d+a e)\right ) \int \frac{1}{d+e x^2} \, dx}{8 d e^3}\\ &=\frac{c x}{e^3}-\frac{\left (c d^2-b d e+a e^2\right ) x}{4 e^3 \left (d+e x^2\right )^2}+\frac{\left (9 c d^2-e (5 b d-a e)\right ) x}{8 d e^3 \left (d+e x^2\right )}-\frac{\left (15 c d^2-e (3 b d+a e)\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{8 d^{3/2} e^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.10437, size = 122, normalized size = 0.98 \[ \frac{x \left (a e^2-5 b d e+9 c d^2\right )}{8 d e^3 \left (d+e x^2\right )}-\frac{x \left (a e^2-b d e+c d^2\right )}{4 e^3 \left (d+e x^2\right )^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (-a e^2-3 b d e+15 c d^2\right )}{8 d^{3/2} e^{7/2}}+\frac{c x}{e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*x^2 + c*x^4))/(d + e*x^2)^3,x]

[Out]

(c*x)/e^3 - ((c*d^2 - b*d*e + a*e^2)*x)/(4*e^3*(d + e*x^2)^2) + ((9*c*d^2 - 5*b*d*e + a*e^2)*x)/(8*d*e^3*(d +
e*x^2)) - ((15*c*d^2 - 3*b*d*e - a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(3/2)*e^(7/2))

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Maple [A]  time = 0.01, size = 179, normalized size = 1.4 \begin{align*}{\frac{cx}{{e}^{3}}}+{\frac{{x}^{3}a}{8\, \left ( e{x}^{2}+d \right ) ^{2}d}}-{\frac{5\,{x}^{3}b}{8\,e \left ( e{x}^{2}+d \right ) ^{2}}}+{\frac{9\,{x}^{3}cd}{8\,{e}^{2} \left ( e{x}^{2}+d \right ) ^{2}}}-{\frac{ax}{8\,e \left ( e{x}^{2}+d \right ) ^{2}}}-{\frac{3\,bdx}{8\,{e}^{2} \left ( e{x}^{2}+d \right ) ^{2}}}+{\frac{7\,c{d}^{2}x}{8\,{e}^{3} \left ( e{x}^{2}+d \right ) ^{2}}}+{\frac{a}{8\,de}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}+{\frac{3\,b}{8\,{e}^{2}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}-{\frac{15\,cd}{8\,{e}^{3}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c*x^4+b*x^2+a)/(e*x^2+d)^3,x)

[Out]

c*x/e^3+1/8/(e*x^2+d)^2/d*x^3*a-5/8/e/(e*x^2+d)^2*x^3*b+9/8/e^2/(e*x^2+d)^2*x^3*c*d-1/8/e/(e*x^2+d)^2*a*x-3/8/
e^2/(e*x^2+d)^2*d*b*x+7/8/e^3/(e*x^2+d)^2*c*d^2*x+1/8/e/d/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))*a+3/8/e^2/(d*e)^
(1/2)*arctan(e*x/(d*e)^(1/2))*b-15/8/e^3*d/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))*c

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c*x^4+b*x^2+a)/(e*x^2+d)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.7991, size = 873, normalized size = 7.04 \begin{align*} \left [\frac{16 \, c d^{2} e^{3} x^{5} + 2 \,{\left (25 \, c d^{3} e^{2} - 5 \, b d^{2} e^{3} + a d e^{4}\right )} x^{3} +{\left (15 \, c d^{4} - 3 \, b d^{3} e - a d^{2} e^{2} +{\left (15 \, c d^{2} e^{2} - 3 \, b d e^{3} - a e^{4}\right )} x^{4} + 2 \,{\left (15 \, c d^{3} e - 3 \, b d^{2} e^{2} - a d e^{3}\right )} x^{2}\right )} \sqrt{-d e} \log \left (\frac{e x^{2} - 2 \, \sqrt{-d e} x - d}{e x^{2} + d}\right ) + 2 \,{\left (15 \, c d^{4} e - 3 \, b d^{3} e^{2} - a d^{2} e^{3}\right )} x}{16 \,{\left (d^{2} e^{6} x^{4} + 2 \, d^{3} e^{5} x^{2} + d^{4} e^{4}\right )}}, \frac{8 \, c d^{2} e^{3} x^{5} +{\left (25 \, c d^{3} e^{2} - 5 \, b d^{2} e^{3} + a d e^{4}\right )} x^{3} -{\left (15 \, c d^{4} - 3 \, b d^{3} e - a d^{2} e^{2} +{\left (15 \, c d^{2} e^{2} - 3 \, b d e^{3} - a e^{4}\right )} x^{4} + 2 \,{\left (15 \, c d^{3} e - 3 \, b d^{2} e^{2} - a d e^{3}\right )} x^{2}\right )} \sqrt{d e} \arctan \left (\frac{\sqrt{d e} x}{d}\right ) +{\left (15 \, c d^{4} e - 3 \, b d^{3} e^{2} - a d^{2} e^{3}\right )} x}{8 \,{\left (d^{2} e^{6} x^{4} + 2 \, d^{3} e^{5} x^{2} + d^{4} e^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c*x^4+b*x^2+a)/(e*x^2+d)^3,x, algorithm="fricas")

[Out]

[1/16*(16*c*d^2*e^3*x^5 + 2*(25*c*d^3*e^2 - 5*b*d^2*e^3 + a*d*e^4)*x^3 + (15*c*d^4 - 3*b*d^3*e - a*d^2*e^2 + (
15*c*d^2*e^2 - 3*b*d*e^3 - a*e^4)*x^4 + 2*(15*c*d^3*e - 3*b*d^2*e^2 - a*d*e^3)*x^2)*sqrt(-d*e)*log((e*x^2 - 2*
sqrt(-d*e)*x - d)/(e*x^2 + d)) + 2*(15*c*d^4*e - 3*b*d^3*e^2 - a*d^2*e^3)*x)/(d^2*e^6*x^4 + 2*d^3*e^5*x^2 + d^
4*e^4), 1/8*(8*c*d^2*e^3*x^5 + (25*c*d^3*e^2 - 5*b*d^2*e^3 + a*d*e^4)*x^3 - (15*c*d^4 - 3*b*d^3*e - a*d^2*e^2
+ (15*c*d^2*e^2 - 3*b*d*e^3 - a*e^4)*x^4 + 2*(15*c*d^3*e - 3*b*d^2*e^2 - a*d*e^3)*x^2)*sqrt(d*e)*arctan(sqrt(d
*e)*x/d) + (15*c*d^4*e - 3*b*d^3*e^2 - a*d^2*e^3)*x)/(d^2*e^6*x^4 + 2*d^3*e^5*x^2 + d^4*e^4)]

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Sympy [A]  time = 3.09714, size = 201, normalized size = 1.62 \begin{align*} \frac{c x}{e^{3}} - \frac{\sqrt{- \frac{1}{d^{3} e^{7}}} \left (a e^{2} + 3 b d e - 15 c d^{2}\right ) \log{\left (- d^{2} e^{3} \sqrt{- \frac{1}{d^{3} e^{7}}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{d^{3} e^{7}}} \left (a e^{2} + 3 b d e - 15 c d^{2}\right ) \log{\left (d^{2} e^{3} \sqrt{- \frac{1}{d^{3} e^{7}}} + x \right )}}{16} + \frac{x^{3} \left (a e^{3} - 5 b d e^{2} + 9 c d^{2} e\right ) + x \left (- a d e^{2} - 3 b d^{2} e + 7 c d^{3}\right )}{8 d^{3} e^{3} + 16 d^{2} e^{4} x^{2} + 8 d e^{5} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(c*x**4+b*x**2+a)/(e*x**2+d)**3,x)

[Out]

c*x/e**3 - sqrt(-1/(d**3*e**7))*(a*e**2 + 3*b*d*e - 15*c*d**2)*log(-d**2*e**3*sqrt(-1/(d**3*e**7)) + x)/16 + s
qrt(-1/(d**3*e**7))*(a*e**2 + 3*b*d*e - 15*c*d**2)*log(d**2*e**3*sqrt(-1/(d**3*e**7)) + x)/16 + (x**3*(a*e**3
- 5*b*d*e**2 + 9*c*d**2*e) + x*(-a*d*e**2 - 3*b*d**2*e + 7*c*d**3))/(8*d**3*e**3 + 16*d**2*e**4*x**2 + 8*d*e**
5*x**4)

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Giac [A]  time = 1.08884, size = 144, normalized size = 1.16 \begin{align*} c x e^{\left (-3\right )} - \frac{{\left (15 \, c d^{2} - 3 \, b d e - a e^{2}\right )} \arctan \left (\frac{x e^{\frac{1}{2}}}{\sqrt{d}}\right ) e^{\left (-\frac{7}{2}\right )}}{8 \, d^{\frac{3}{2}}} + \frac{{\left (9 \, c d^{2} x^{3} e - 5 \, b d x^{3} e^{2} + 7 \, c d^{3} x + a x^{3} e^{3} - 3 \, b d^{2} x e - a d x e^{2}\right )} e^{\left (-3\right )}}{8 \,{\left (x^{2} e + d\right )}^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c*x^4+b*x^2+a)/(e*x^2+d)^3,x, algorithm="giac")

[Out]

c*x*e^(-3) - 1/8*(15*c*d^2 - 3*b*d*e - a*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-7/2)/d^(3/2) + 1/8*(9*c*d^2*x^3*e
- 5*b*d*x^3*e^2 + 7*c*d^3*x + a*x^3*e^3 - 3*b*d^2*x*e - a*d*x*e^2)*e^(-3)/((x^2*e + d)^2*d)